1
$\begingroup$

Let the process $$X_t:=\cos(B_{1,t}B_{2,t}),$$ where $(B_1, B_2)$ is a bi-dimensional correlated Brownian motion, be given. Is the following the correct Ito representation of its stochastic differential? $$dX_t = \\-\frac12 \left\{\rho^{1,1}B_{2,t}^2\cos{(B_{1,t}B_{2,t})} + \rho^{2,2}B_{1,t}^2\cos{(B_{1,t}B_{2,t})} \\+ \rho^{2,1}[\sin{(B_{1,t}B_{2,t})} + B_{1,t}B_{2,t}\cos{(B_{1,t}B_{2,t})}]\right\}dt \\- B_{2,t}\sin{(B_{1,t}B_{2,t})}dB_{1,t}-B_{1,t}\sin{(B_{1,t}B_{2,t})}dB_{2,t}$$

$\endgroup$

1 Answer 1

2
$\begingroup$

Let $Y_t:=B_{1,t}B_{2,t}$, so that $X_t=\cos(Y_t)$ and $$ dX_t=-\sin(Y_t)dY_t -{1\over 2}\cos(Y_t)d\langle Y\rangle_t. $$ To make this more explicit, use $$ dY_t=B_{1,t} dB_{2,t}+B_{2,t} dB_{1,t}+\rho^{2,1} dt $$ (if I guess correctly at what you mean by $\rho^{2,1}$), and $$ d\langle Y\rangle_t = B_{1,t}^2\rho^{2,2} dt +B_{2,t}^2\rho^{1,1} dt + 2B_{1,t}B_{2,t}\rho^{2,1} dt. $$ It looks to me like your formula is correct except your $-{1\over 2}\rho^{2,1}\sin(B_{1,t}B_{2,t}) dt$ term shouldn't have a ${1\over 2}$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .